Question

在△ABC中,∠C=90°,AC=CB,AD是∠BAC的平分线,DE⊥AB于E,若AB=10cm,求△BDE的周长

Answer 1

AC=AB*cos∠BAC=50√2

CD=AC*tg∠CAD

=AC*tg(1/2∠BAC)

=AC*(1-cos∠BAC)/sin∠BAC

=5√2*(√2-1)

=10-5√2

BD=BC-CD

=5√2-(10-5√2)

=10(√2-1)

BE=DE

=BD*√2/2

=5(2-√2)

那么周长就是

L=BD+BE+DE

=10（√2-1）+2*5(2-√2)=10.